Multiple choice question of indefinite integral, $\int \frac{x + 9}{x^3 + 9x} dx$. If $\int \frac{x + 9}{x^3 + 9x} dx = k\arctan(mx) + n\ln (x) + p \ln (x^2 + 9) + c$, then $(m+n)/(k+p) = $
(A) 6
(B) -8
(C) -3
(D) 4  
I tried solving it by differentiating the R.H.S. but couldn't arrive at the answer.
 A: The derivative of the right hand side is
$$
\frac{km}{1+m^2x^2}+\frac{n}{x}+\frac{2px}{x^2+9}
$$
which should equal
$$
\frac{x + 9}{x^3 + 9x}
$$
Thus you need $m=1/3$ and then you can go on.
A: \begin{align}
\int \frac{x + 9}{x^3 + 9x} dx &= \int \frac{x}{(x^2 + 9)x} dx + \int \frac{9}{(x^2 + 9)x} dx 
\\
\\
&= \int \frac{dx}{x^2 + 9}  + \int \left(\frac{A}{x} + \frac{Bx+C}{x^2 + 9}\right) dx 
\\
\\
&= \int \frac{dx}{x^2 + 9}  + \int \left(\frac{1}{x} + \frac{-x}{x^2 + 9}\right) dx 
\\
\\
&= \frac{1}{9}\int \frac{dx}{\left(x/3\right)^2 + 1}  + \int \frac{dx}{x}  - \int\frac{x}{x^2 + 9}dx 
\\
\\
&= \frac{1}{3}\arctan(x/3)  + \ln x  - \frac{1}{2}\ln(x^2+9) + c
\end{align}
$$\int \frac{x + 9}{x^3 + 9x} dx = k\arctan(mx) + n\ln (x) + p \ln (x^2 + 9) + c$$
$$\frac{m+n}{k+p} = \frac{1/3+1}{1/3-1/2}=\frac{4/3}{-1/6}=-8$$
A: Try to separate the denominator like this
$$\frac{x+9}{x^3+9x}=\frac{x+9}{x(x^2+9)}$$
$$\qquad\qquad\ \,=\frac Ax+\frac{Bx+C}{x^2+9}$$
Then construct an equality for $A,B,C$ and we have
$$Ax^2+9A+Bx^2+Cx=x+9$$
Here we get $A=1,\ B=-1,\ C=1$, and that gives us
$$\int\frac{x+9}{x^3+9x}\,dx\ =\ \int\frac1x+\frac1{x^2+9}-\frac x{x^2+9}\,dx$$
Could you continue with this?
$----------$
Note that
\begin{align*}
&\int\frac{x}{x^2+9}\,dx\\
=\ \frac12&\int\frac{du}{u+9}\qquad(u=x^2,\ du=2x\,dx)\\
=\ \frac12&\ln(u+9)+c\\
=\ \frac12&\ln(x^2+9)+c
\end{align*}
